1 Lagrangian for a continuous system Let’s start with an example from mechanics to get the big idea. The physical system of interest is a string of length and mass per unit length fixed at both ends, and under tension Choose −axis along the unperturbed string, and −axis perpendicular to it. When the string is vibrating, its kinetic energy is: = Z 0 1 2 = 2 Z 0 1 2 µ ¶2 = Z 0 1 (̇)2 2 To get the potential energy, we use the method of virtual work. The net force on a string segment has components: = cos 1 − cos 2 ' 0 and = sin 1 − sin 2 ≈ tan 1 − tan 2 = = à 2 2 ¯ ¯ ! ¯¯ ¯¯ − ¯+ ¯ Then the virtual work is = Z = 0 Z 0 2 2 Now integrate by parts, and make use of the fixed end condition: " # Z Z h 0 2i 0 0 0 = |0 − ( ) = − ( ) 2 0 0 R then · = − and = − · Here Then if = −∇ =− Z = 2 Z 2 ( 0 ) 0 The Lagrangian for the string is: = − = Z 0 i 1h 2 2 (̇) − ( 0 ) 2 where i 1h 2 2 (̇) − ( 0 ) 2 is the Lagrangian density for the string. L= 1 (1) The action is = Z = Z 2 1 Z 0 i 1h 2 2 (̇) − ( 0 ) 2 Taking the variation of the action, we get ¸ Z 2 Z ∙ L L 0 L = ̇ + 0 + ̇ 1 0 Integrating by parts gives: Z 2 Z = 1 0 ∙ ¸ L L L + − − ̇ 0 Thus for the action to be an extremum, we need − L L L + − =0 0 ̇ (2) Using equation (1), we find: (2̇) + (−2 0 ) − 0 = 0 or ·· − 00 = 0 which is the wave equation for the string. An alternative approach is to write the string displacement as a sum over normal modes: X = () sin Then the Lagrangian density (1) is L= XX ̇ ̇ sin 2 sin − 2 cos cos and then the Lagrangian is = Z L When we integrate over the only terms that survive are those with = µ ¶ X 2 2 = (̇ ) − 2 2 2 (3) 2 The mode amplitudes act as the generalized coordinates for the string. Then Lagrange’s equations are 2 ·· − = − 2 2 = 0 ̇ which is the harmonic oscillator equation with frequency = () 2 p 2 Lagrangian for the electromagnetic field Now we want to do a similar treatment for the EM field. We want a Lagrangian density such that the action Z = L4 is a Lorentz invariant, and where L is a function of the fields. The "obvious" invariant to try is Lguess = (Recall this is proportional to 2 − 2 an "energy-like" thing.) Here the components of the potential are the "normal modes" — they behave like the in the previous section. Then Lagrange’s equations (2) are: L L ³ ´− =0 (4) To evaluate this, note that ¡ ¢ ¡ ¢ Lguess = − − and so Lguess ( ) ´ ³ ´ ³ ¡ ¢ ¡ ¢ − − − − + ¡ ¢ ¡ ¢ = ( − ) − + − ( − ) = ( − ) − ( − ) + ( − ) − ( − ) = 4 ( − ) = 4 = while Lguess =0 So equations (4) become = 0 which are Maxwell’s equations in the absence of sources. We can fix up the Lagrangian by adding the interaction term 1 Thus L=− 1 1 − 16 With this Lagrangian density 1 L = − and Lagrange’s equations become − 1 1 + = 0 4 3 or 4 which are the two Maxwell equations that include sources. = 3 The Hamiltonian Now we form the Hamiltonian. First let’s look at the string. Using equation (3): X L ̇ − L ̇ µ ¶ 2 X 2 2 2 (̇ ) − (̇ ) − 2 2 2 = 2 2 X X = 2 (̇ )2 + 2 2 2 = 2 = where is the total (kinetic plus potential) energy per mode. By analogy, we get for the EM field system without sources L − L ( ) µ ¶ 1 1 = − − − 4 16 ¶ µ 1 1 = + 4 4 = This tensor is not symmetric, because the first term contains only one half of the field tensor: rather than The conservation laws require that the energy tensor be symmetric, so we have to modify the result. 4 The energy-momentum tensor Recall that the field energy density (non-relativistic) is Poynting theorem may be written 1 8 ¡ 2 ¢ + 2 and the ¢ 1 ¡ 2 · × + · =0 + 2 + ∇ 8 4 4 (5) We’d like to express this result in covariant form. We obviously need something quadratic in the fields. For example: ⎞⎛ ⎞ ⎛ 0 − − − 0 ⎜ ⎜ 0 − ⎟ 0 − ⎟ ⎟ ⎜ ⎟ = ⎜ ⎝ − 0 ⎠ ⎝ 0 − ⎠ − 0 − 0 ⎛ 2 + 2 + 2 − − − ⎜ − −2 + 2 + 2 − − − − = ⎜ ⎝ − − − −2 + 2 + 2 − − − − − − − −2 + 2 + 2 ³ ´ ³ ´ ³ ´ ⎞ ⎛ × × × 2 ´ ⎟ ⎜ ³ ⎟ ⎜ 2 2 2 − + + − − − − × ⎜ ⎟ ⎟ ⎜ ³ ´ = ⎜ ⎟ × − − −2 + 2 + 2 − − ⎟ ⎜ ⎠ ⎝ ³ ´ × − − − − −2 + 2 + 2 ⎞ ⎟ ⎟ ⎠ Now we’d like the (0,0) component to be the ¡ energy¢ density. We can get that if we add the tensor 14 = 12 2 − 2 Then ¾ ½ 1 1 + Θ = 4 4 ³ ´ ³ ´ ³ ´ ⎛ 2 2 × × × 2 + − 2 ´ ⎜ ³ ⎜ 2 2 2 2 − 2 − + + − 2 − − − − × 1 ⎜ ⎜ ³ ´ = 2 2 ⎜ 2 2 2 − × 4 ⎜ − − − + + − 2 − − ⎝ ³ ´ 2 2 × − − − − − 2 + 2 + 2 − − 2 ³ ´ ³ ´ ³ ´ ⎛ 2 + 2 ⎞ × × × 2 ´ ⎜ ³ ⎟ ⎜ ⎟ 2 + 2 2 2 × − − − − − − ⎟ 1 ⎜ 2 ⎜ ⎟ ³ ´ = 2 2 ⎜ ⎟ + 2 2 × 4 ⎜ − − − − − − ⎟ 2 ⎝ ³ ⎠ ´ 2 2 + 2 2 × − − − − − − 2 Then 0 Θ 2 + 2 = +∇ 8 Ã × 4 ! = 1 0 which is part of equation (5). On the right hand side we need 1 · Thus we have the relation Θ0 = 5 1 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ and so we guess that the full set of conservation laws are given by: Θ = 1 I leave it to you to show that the = components give momentum conservation (Jackson equation 6.122). 5 Angular momentum Cross products are not proper vectors. They are pseudo-vectors because they do not transform properly under reflections. Thus it is usually better to express = × ) as antisymmetric quantities such as angular momentum of a particle ( tensors. For example the tensor X = ( − ) particles has three independent compomemts: the components of the vector Extending this idea, let’s look at the tensor X ¡ ¢ = − particles The 3×3 spacelike part is the tensor and thus represents the angular momentum of the system. In addition: ´ X³ 0 = − where is the energy of the particle. Conservation of angular momentum for the system is expressed as = constant. Thus we conjecture that the full conservation law is = constant. (Or equivalently = 0) This gives for the ( 0) component: ´ X³ 0 = − = constant P Now if we divide through by we get P P P = 2 P The term on the left hand side is the position of the center of mass, P = P while the term on the right hand side is the CM velocity times P = P 6 an eminently sensible result. To get the equivalent result for the EM field we form the tensor = Θ − Θ and then the conservation laws should be given by: = 0 Taking = 0 gives the CM motion as above. 6 The Darwin Lagrangian The analysis above is for source-free fields. We might attempt to add the freeparticle Lagrangian to get a complete description of the particle-plus-field system, but this approach fails because of retardation effects. (The fields propagate at the speed of light.) We can calculate a complete Lagrangian in a single reference frame, inlcuding relativistic effects up to order 2 = ()2 Let’s start with a 2-particle system. Both particles produce fields and both can move under the influence of those fields. The interaction term for charge 1 interacting with the fields due to 2 is µ ¶ ³ ´ 1 1 1 = ) − (6) = 1 ( · 1 2 1 2 2 2 2 If we now work in a single reference frame and use the coordinate time rather than proper time as our time variable, we should drop the factor We want to evaluate this expression to second order in If we work in Coulomb gauge, to first order since it appears the potential 2 = 2 is exact. We only need in combination with 1 This means we can ignore retardation effects. Then: 2 ' 1 Z 0 | − 0 | where the transverse current is Therefore Z 0 1 ∇ · 22 (0 − 2 ) 0 = − = 22 ( − 2 ) − ∇ 4 | − 0 | 2 2 · ( − 2 ) = 22 ( − 2 ) − ∇ 4 | − 2 |3 = 2 2 − 2 4 Z 0 1 0 2 · ( − 2 ) 0 ∇ 3 | − 0 | |0 − 2 | 7 Let’s look at the integral. First make a change of origin. Let = 0 − 2 and with = − 2 we get Z Z 0 1 1 2 · 3 0 2 · ( − 2 ) 0 = ∇ ∇ 3 3 0 0 | − | | − | | − 2 | ¯ µ ¶ Z ¯ 1 2 · 3 1 2 · ¯ − ∇ = | − | 3 ¯ at ∞ − 3 Z 1 2 · ̂ 3 = ∇ | − | 2 Z X ∞ () 2 · ̂ = ∇ +1 =0 where we have put the polar axis for along Now 2 · ̂ = 2 · (̂ cos + sin (̂ cos + ̂ sin )) Integration over renders the − and −components zero. Next we make use of the orthogonality of the () noting that cos = 1 () Only = 1 survives the integration over We obtain: Z ∞ 2 integral = 2∇ 2 · ̂ 2 3 0 µZ ¶ Z ∞ 4 = + ∇ 2 · ̂ 3 2 2 µ 0 ¶ µ ¶ 4 1 = ∇ 2 · ̂ + 1 = 2∇ 2 · 3 2 And thus µ ¶ 2 2 2 = − 2∇ 2 · 4 ¶¸ ∙ µ 2 2 1 2 2 · = − − 2 ̂ 2 2 = [2 + (2 · ̂) ̂] 2 Then the interaction term for 2 particles (equation 6 with the dropped) is: à ! ¶ µ 2 1 · 2 1 2 1 2 − = 1 − · [2 + (2 · ̂) ̂] 2 ¾ ½ 1 2 = 1 1 − 2 [1 · 2 + (2 · ̂) (1 · ̂)] 2 2 Adding this term to the kinetic energy (Lagrangian notes pg 5), we have the Darwin Lagrangian for a collection of charged particles: r ¾ ½ 1X 2 X 1 2 = − 1 − 2 − 1 − 2 [ · + ( · ̂) ( · ̂)] 2 2 8 To be consistent, we should evaluate the first term to second order in i.e. ¢12 ¡ ¢ 4 ¡ 2 ' 1 − 12 2 + 12 − 12 2 1 − 2 2 4 Finally, dropping the constant leading term which is irrelevant, we have ¾ X ½ 1X 1 X 1 2 4 + 2 − 1 − 2 [ · + ( · ̂) ( · ̂)] = 2 8 2 correct to second order in 9
© Copyright 2026 Paperzz